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230
ANALYTICAL MECHANICS
Therefore
and
v = ~ g sm a. o
Thus both the linear and the angular accelerations of the cylinder are constant.   Therefore the equations of the motion are the following:
o
- gt sin a,
o
s = vQt + - gtf2 sin a,
o
2;2 = y2+-o  -Tg
co0 + ;j— <7 sin or,
—
o a
4
7T~
oa
sn a
2. A wheel moves down an inclined groove with its axle rolling along the groove without slipping.   Discuss the motion. Let a — the radius of the axle. b = the radius of the wheel. m'= the mass of that part of the axle which projects out from the wheel. m = the mass of the rest of the
moving system. M = m + mf.
Suppose the wheel to be a solid disk with a thickness equal to half the total length of the axle. Then if both the wheel and the axle are of the same
FIG. 116.
material the relation — = —
m
Therefore
m'
; M   and m =
TORQUE METHOD.—Considering the moments about the element of contact we obtain the following for the torque equation:
I(a = Mga sin a, (Ic+Ma?) o) — Mga sin a, (Ic + Maz) v = Mga2 sin a, where Ic denotes the moment of inertia of the moving system about its own axis.
M.
ENERGY METHOD. — Supposing the wheel to start from rest we obtain
+ $ Jcco2 = Mgs sin a,
/c2   = Mussina,
(Ic + Ma*) v2 = Mga*s sin a, (Ic + Ma2) vv = Mga2s sin a, (Ic + Ma2) v = M#a2 sin a.